Problem: Let $f(x)=\begin{cases} \ln(x)&\text{for }0<x\leq2 \\\\ x^2\ln(2)&\text{for }x>2 \end{cases}$ Is $f$ continuous at $x=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
For $f$ to be continuous at $x=2$, we need $\lim_{x\to 2}f(x)$ and $f(2)$ to exist and be equal. Since $2\leq 2$, the rule that applies to $x=2$ is $\ln(x)$. So $f(2)=\ln(2)$. Now let's analyze $\lim_{x\to 2}f(x)$. Finding $\lim_{x\to 2^{ +}}f(x)$ For $x$ -values larger than $2$, the appropriate rule for $f(x)$ is $x^2\ln(2)$. Since $x^2\ln(2)$ is continuous for all real numbers, any limit is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 2^{ +}}f(x) \\\\ &=\lim_{x\to 2^{ +}}[x^2\ln(2)] \gray{x^2\ln(2)\text{ is the rule for }x>2} \\\\ &=(2)^2\ln(2) \gray{x^2\ln(2)\text{ is continuous at }x=2} \\\\ &=\ln(16) \end{aligned}$ Finding $\lim_{x\to 2^{ -}}f(x)$ For $x$ -values smaller than $2$, the appropriate rule for $f(x)$ is $\ln(x)$. Since $\ln(x)$ is continuous for $0<x<2$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 2^{ -}}f(x) \\\\ &=\lim_{x\to 2^{ -}}[\ln(x)] \gray{\ln(x)\text{ is the rule for }x<2} \\\\ &=\ln(2) \gray{\ln(x)\text{ is continuous at }x=2} \end{aligned}$ Conclusion We found that $\lim_{x\to 2^{ +}}f(x)=\ln(16)$ and $\lim_{x\to 2^{ -}}f(x)=\ln(2)$. Since the one-sided limits aren't equal, $\lim_{x\to 2}f(x)$ doesn't exist and $f$ isn't continuous at $x=2$.